Chapter 5 Normal Probability Distributions Normal Probability Distributions Chapter 5 5.1...

1

Normal Probability

Distributions

Chapter 5

§ 5.1

Introduction to Normal

Distributions and the

Standard Distribution

Larson & Farber, Elementary Statistics: Picturing the World, 3e 3

Properties of Normal Distributions

A continuous random variable has an infinite number of possible

values that can be represented by an interval on the number line.

Hours spent studying in a day

0 63 9 1512 18 2421

The time spent studying

can be any number

between 0 and 24.

The probability distribution of a continuous random variable is

called a continuous probability distribution.

2



The most important probability distribution in statistics is the

normal distribution.

A normal distribution is a continuous probability distribution for

a random variable, x. The graph of a normal distribution is called

the normal curve.

Normal curve

x



Properties of a Normal Distribution

1. The mean, median, and mode are equal.

2. The normal curve is bell-shaped and symmetric about the mean.

3. The total area under the curve is equal to one.

4. The normal curve approaches, but never touches the x-axis as it

extends farther and farther away from the mean.

5. Between µ − σ and µ + σ (in the center of the curve), the graph

curves downward. The graph curves upward to the left of µ − σ

and to the right of µ + σ. The points at which the curve changes

from curving upward to curving downward are called the

inflection points.



µ − 3σ µ + σµ − 2σ µ − σ µ µ + 2σ µ + 3σ

Inflection points

Total area = 1

If x is a continuous random variable having a normal distribution

with mean µ and standard deviation σ, you can graph a normal

curve with the equation2 2( ) 21

=2

x µ σy eσ π

- - . = 2.178 = 3.14 e π

x

3


Means and Standard Deviations

A normal distribution can have any mean and any

positive standard deviation.

Mean: µ = 3.5

Standard deviation:

σ ≈≈≈≈ 1.3

Mean: µ = 6

Standard deviation:

σ ≈≈≈≈ 1.9

The mean gives the

location of the line

of symmetry.

The standard deviation describes the spread of the data.

Inflection

pointsInflection

points

3 61 542x

3 61 542 97 11108x


Means and Standard Deviations

Example:

1. Which curve has the greater mean?

2. Which curve has the greater standard deviation?

The line of symmetry of curve A occurs at x = 5. The line of symmetry of curve

B occurs at x = 9. Curve B has the greater mean.

Curve B is more spread out than curve A, so curve B has the greater standard

deviation.

31 5 97 11 13

AB

x


Interpreting Graphs

Example:

The heights of fully grown magnolia bushes are normally

distributed. The curve represents the distribution. What is the

mean height of a fully grown magnolia bush? Estimate the

standard deviation.

The heights of the magnolia bushes are normally distributed with

a mean height of about 8 feet and a standard deviation of about

0.7 feet.

µ = 8

The inflection points are one standard

deviation away from the mean.

σ ≈ 0.7

6 87 9 10Height (in feet)

x

4


−3 1−2 −1 0 2 3z

The Standard Normal Distribution

The standard normal distribution is a normal distribution with a

mean of 0 and a standard deviation of 1.

Any value can be transformed into a z-score by using the formula

The horizontal scale

corresponds to z-scores.

- -Value Mean= = .Standa rd deviat ion

x µz

σ


The Standard Normal Distribution

If each data value of a normally distributed random variable x is

transformed into a z-score, the result will be the standard normal

distribution.

After the formula is used to transform an x-value into a z-score,

the Standard Normal Table in Appendix B is used to find the

cumulative area under the curve.

The area that falls in the interval under the

nonstandard normal curve (the x-values) is the

same as the area under the standard normal

curve (within the corresponding z-

boundaries).

−3 1−2 −1 0 2 3

z


The Standard Normal Table

Properties of the Standard Normal Distribution

1. The cumulative area is close to 0 for z-scores close to z = −3.49.

2. The cumulative area increases as the z-scores increase.

3. The cumulative area for z = 0 is 0.5000.

4. The cumulative area is close to 1 for z-scores close to z = 3.49

z = −3.49

Area is close to 0.

z = 0Area is 0.5000.

z = 3.49

Area is close to 1.z

−3 1−2 −1 0 2 3

5



Example:

Find the cumulative area that corresponds to a z-score of 2.71.

.6141 .6103 .6064 .6026 .5987 .5948 .5910 .5871 .5832 .5793 0.2

.5753 .5714 .5675 .5636 .5596 .5557 .5517 .5478 .5438 .5398 0.1

.5359 .5319 .5279 .5239 .5199 .5160 .5120 .5080 .5040 .5000 0.0

.09 .08 .07 .06 .05 .04 .03 .02 .01 .00 z

.9981 .9980 .9979 .9979 .9978 .9977 .9977 .9976 .9975 .9974 2.8

.9974 .9973 .9972 .9971 .9970 .9969 .9968 .9967 .9966 .9965 2.7

.9964 .9963 .9962 .9961 .9960 .9959 .9957 .9956 .9955 .9953 2.6

Find the area by finding 2.7 in the left hand column, and then

moving across the row to the column under 0.01.

The area to the left of z = 2.71 is 0.9966.

Appendix B: Standard Normal Table



Example:

Find the cumulative area that corresponds to a z-score of −0.25.

.0005.0005.0005.0004.0004.0004.0004.0004.0004.0003−−−−3.3

.0003.0003.0003.0003.0003.0003.0003.0003.0003.0002−−−−3.4

.00.01.02.03.04.05.06.07.08 .09 z

Find the area by finding −0.2 in the left hand column, and then

moving across the row to the column under 0.05.

The area to the left of z = −0.25 is 0.4013

.5000.4960.4920.4880.4840.4801.4761.4724.4681.4641−−−−0.0

.4602.4562.4522.4483.4443.4404.4364.4325.4286.4247 −−−−0.1

.4207.4168.4129.4090.4052.4013.3974.3936.3897.3859−−−−0.2

.3821.3783.3745.3707.3669.3632.3594.3557.3520.3483 −−−−0.3



Guidelines for Finding Areas

Finding Areas Under the Standard Normal Curve

1. Sketch the standard normal curve and shade the appropriate area under the curve.

2. Find the area by following the directions for each case shown.

a. To find the area to the left of z, find the area that corresponds to z in the Standard Normal Table.

1. Use the table to find the

area for the z-score.

2. The area to the left

of z = 1.23 is

0.8907.

1.230

z

6




b. To find the area to the right of z, use the Standard Normal

Table to find the area that corresponds to z. Then subtract

the area from 1.

3. Subtract to find the area to the

right of z = 1.23: 1 −0.8907 = 0.1093.

1. Use the table to find the

area for the z-score.

2. The area to the

left of z = 1.23 is

0.8907.

1.230

z



c. To find the area between two z-scores, find the area

corresponding to each z-score in the Standard Normal

Table. Then subtract the smaller area from the larger area.


4. Subtract to find the area of the

region between the two z-scores:

0.8907 − 0.2266 = 0.6641.

1. Use the table to find the area for the z-

score.

3. The area to the left of

z = −0.75 is 0.2266.

2. The area to the

left of z = 1.23 is

0.8907.

1.230

z

−0.75



Example:

Find the area under the standard normal curve to

the left of z = −2.33.

From the Standard Normal Table, the area is equal to

0.0099.

Always draw

the curve!

−2.33 0

z

7



Example:

Find the area under the standard normal curve to the

right of z = 0.94.

From the Standard Normal Table, the area is equal to 0.1736.

Always draw the

curve!0.8264

1 − 0.8264 = 0.1736

0.940

z



Example:

Find the area under the standard normal curve

between z = −1.98 and z = 1.07.

From the Standard Normal Table, the area is equal to 0.8338.

Always draw

the curve!

0.8577 − 0.0239 = 0.8338

0.8577

0.0239

1.070

z

−1.98

§ 5.2

Normal Distributions:

Finding Probabilities

8


Probability and Normal Distributions

If a random variable, x, is normally distributed, you can

find the probability that x will fall in a given interval by

calculating the area under the normal curve for that

interval.

P(x < 15)µ = 10

σ = 5

15µ =10x



Same area

P(x < 15) = P(z < 1) = Shaded area under the curve

= 0.8413

15µ =10

P(x < 15)

µ = 10

σ = 5

Normal Distribution

x

1µ =0

µ = 0

σ = 1

Standard Normal Distribution

z

P(z < 1)


Example:

The average on a statistics test was 78 with a standard deviation of

8. If the test scores are normally distributed, find the probability

that a student receives a test score less than 90.


P(x < 90) = P(z < 1.5) = 0.9332

-=

90 -78=

8x µ

zσ

= 1.5

The probability that a student

receives a test score less than

90 is 0.9332.

µ =0

z

?1.5

90µ =78

P(x < 90)

µ = 78

σ = 8

x

9


Example:



that a student receives a test score greater than than 85.


P(x > 85) = P(z > 0.88) = 1 − P(z < 0.88) = 1 − 0.8106 = 0.1894

85 -78= =

8x - µ

zσ

≈= 0.875 0.88


receives a test score greater

than 85 is 0.1894.

µ =0

z

?0.88

85µ =78

P(x > 85)

µ = 78

σ = 8

x


Example:



that a student receives a test score between 60 and 80.


P(60 < x < 80) = P(−2.25 < z < 0.25) = P(z < 0.25) − P(z < −2.25)

- -1

60 78= =

8x µ

zσ

-= 2.25


receives a test score between

60 and 80 is 0.5865.

2

- -=

80 78=

8x µ

zσ

= 0.25

µ =0

z

?? 0.25−2.25

= 0.5987 − 0.0122 = 0.5865

60 80µ =78

P(60 < x < 80)

µ = 78

σ = 8

x

§ 5.3

Normal Distributions:

Finding Values

10


Finding z-Scores

Example:

Find the z-score that corresponds to a cumulative area of 0.9973.

.6141 .6103 .6064 .6026 .5987 .5948 .5910 .5871 .5832 .5793 0.2

.5753 .5714 .5675 .5636 .5596 .5557 .5517 .5478 .5438 .5398 0.1

.5359 .5319 .5279 .5239 .5199 .5160 .5120 .5080 .5040 .5000 0.0

.09 .08 .07 .06 .05 .04 .03 .02 .01 .00 z

.9981 .9980 .9979 .9979 .9978 .9977 .9977 .9976 .9975 .9974 2.8

.9974 .9973 .9972 .9971 .9970 .9969 .9968 .9967 .9966 .9965 2.7

.9964 .9963 .9962 .9961 .9960 .9959 .9957 .9956 .9955 .9953 2.6

Find the z-score by locating 0.9973 in the body of the Standard Normal

Table. The values at the beginning of the corresponding row and at the

top of the column give the z-score.

The z-score is 2.78.


2.7

.08


Finding z-Scores

Example:

Find the z-score that corresponds to a cumulative area of 0.4170.

.0005.0005.0005.0004.0004.0004.0004.0004.0004.0003−−−−0.2

.0003.0003.0003.0003.0003.0003.0003.0003.0003.0002−−−−3.4

.00.01.02.03.04.05.06.07.08 .09 z

Find the z-score by locating 0.4170 in the body of the Standard Normal

Table. Use the value closest to 0.4170.

.5000.4960.4920.4880.4840.4801.4761.4724.4681.4641−−−−0.0

.4602.4562.4522.4483.4443.4404.4364.4325.4286.4247 −−−−0.1

.4207.4168.4129.4090.4052.4013.3974.3936.3897.3859−−−−0.2

.3821.3783.3745.3707.3669.3632.3594.3557.3520.3483 −−−−0.3


Use the

closest

area.

The z-score is −0.21.

−−−−0.2

.01


Finding a z-Score Given a Percentile

Example:

Find the z-score that corresponds to P75.

The z-score that corresponds to P75 is the same z-score that corresponds

to an area of 0.75.

The z-score is 0.67.

?µ =0

z

0.67

Area = 0.75

11


Transforming a z-Score to an x-Score

To transform a standard z-score to a data value, x, in a given

population, use the formula

Example:

The monthly electric bills in a city are normally distributed with a mean

of $120 and a standard deviation of $16. Find the x-value

corresponding to a z-score of 1.60.

=x µ + zσ.

=x µ + zσ= 120 +1.60(16)

= 145.6

We can conclude that an electric bill of $145.60 is 1.6 standard

deviations above the mean.


Finding a Specific Data Value

Example:

The weights of bags of chips for a vending machine are normally

distributed with a mean of 1.25 ounces and a standard deviation of 0.1

ounce. Bags that have weights in the lower 8% are too light and will

not work in the machine. What is the least a bag of chips can weigh and

still work in the machine?

=x µ + zσ

The least a bag can weigh and still work in the machine is 1.11 ounces.

? 0

z

8%

P(z < ?) = 0.08

P(z < −1.41) = 0.08

−−−−1.41

1.25

x

?1.25 ( 1.41)0.1= + −

= 1.111.11

§ 5.4

Sampling Distributions

and the Central Limit

Theorem

12


Population

Sample


A sampling distribution is the probability distribution of a sample

statistic that is formed when samples of size n are repeatedly taken

from a population.

Sample

Sample

Sample Sample

Sample

Sample

Sample

Sample

Sample



If the sample statistic is the sample mean, then the distribution is

the sampling distribution of sample means.

Sample 1

1xSample 4

4x

Sample 3

3x Sample 6

6x

The sampling distribution consists of the values of the sample

means, 1 2 3 4 5 6, , , , , . x x x x x x

Sample 2

2xSample 5

5x


Properties of Sampling Distributions

Properties of Sampling Distributions of Sample Means

1. The mean of the sample means, is equal to the population mean.

2. The standard deviation of the sample means, is equal to the population

standard deviation, divided by the square root of n.

The standard deviation of the sampling distribution of the sample means is

called the standard error of the mean.

,x

µ

xµ = µ

,x

σ

,σ

x

σσ =

n

13


Sampling Distribution of Sample Means

Example:The population values {5, 10, 15, 20} are written on slips of paper and

put in a hat. Two slips are randomly selected, with replacement.

a. Find the mean, standard deviation, and variance of the

population.

Continued.

= 12.5µ

= 5.59σ

2 = 31.25σ

Population

5

10

15

20



Example continued:

The population values {5, 10, 15, 20} are written on slips of paper and


b. Graph the probability histogram for the population values.

Continued.

This uniform distribution

shows that all values have the

same probability of being

selected.

Population values

Probability

0.25

5 10 15 20

x

P(x) Probability Histogram of

Population of x



Example continued:



c. List all the possible samples of size n = 2 and calculate the

mean of each.

1510, 20

12.510, 15

1010, 10

7.510, 5

12.55, 20

105, 15

7.55, 10

55, 5

Sample mean, Sample x

2020, 20

17.520, 15

1520, 10

12.520, 5

17.515, 20

1515, 15

12.515, 10

1015, 5

Sample mean, Sample x

Continued.

These means

form the

sampling

distribution of

the sample

means.

14



Example continued:



d. Create the probability distribution of the sample means.

Probability Distribution of

Sample Means

0.0625120

0.1250217.5

0.1875315

0.2500412.5

0.1875310

0.125027.5

0.062515

x f Probability



Example continued:



e. Graph the probability histogram for the sampling distribution.

The shape of the graph is

symmetric and bell shaped. It

approximates a normal

distribution.

Sample mean

Probability

0.25

P(x) Probability Histogram of

Sampling Distribution

0.20

0.15

0.10

0.05

17.5 201512.5107.55

x


the sample means will have a normal distribution.

The Central Limit Theorem

If a sample of size n ≥ 30 is taken from a population with any type of

distribution that has a mean = µ and standard deviation = σ,

xµ

xµ

µ

x

xx

x

xxxx x

xxx

x

15



If the population itself is normally distributed, with mean = µ

and standard deviation = σ,

the sample means will have a normal distribution for any

sample size n.

µx

µ

x

xx

x

xxxx x

xxx

x



In either case, the sampling distribution of sample means has a mean

equal to the population mean.

=xµ µ

=x

σσ

n

Mean of the

sample means

Standard deviation of the

sample means

The sampling distribution of sample means has a standard deviation

equal to the population standard deviation divided by the square root

of n.

This is also called the standard

error of the mean.


The Mean and Standard Error

Example:

The heights of fully grown magnolia bushes have a mean height of

8 feet and a standard deviation of 0.7 feet. 38 bushes are randomly

selected from the population, and the mean of each sample is

determined. Find the mean and standard error of the mean of the

sampling distribution.

=x

µ µMean

Standard deviation

(standard error)

=x

σσ

n= 8

0.7=

38= 0.11

Continued.

16


Interpreting the Central Limit Theorem

Example continued:

The heights of fully grown magnolia bushes have a mean height

of 8 feet and a standard deviation of 0.7 feet. 38 bushes are

randomly selected from the population, and the mean of each

sample is determined.

From the Central Limit Theorem,

because the sample size is greater than

30, the sampling distribution can be

approximated by the normal

distribution.

The mean of the sampling distribution is 8 feet ,and the standard

error of the sampling distribution is 0.11 feet.

x

8 8.47.6

= 8x

µ = 0.11xσ



Example:

The heights of fully grown magnolia bushes have a mean height

of 8 feet and a standard deviation of 0.7 feet. 38 bushes are

randomly selected from the population, and the mean of each

sample is determined.

Find the probability that the mean

height of the 38 bushes is less than

7.8 feet.

The mean of the sampling distribution is 8

feet, and the standard error of the sampling

distribution is 0.11 feet.

7.8

x

8.47.6 8

Continued.

=8x

µ

=0.11xσ

=38n


P ( < 7.8) = P (z < ____ )?x −−−−1.82


Example continued:

Find the probability that the mean height of the 38 bushes is less

than 7.8 feet.

−= x

x

x µz

σ

−7.8 8=

0.117.8

x

8.47.6 8

= 8x

µ

= 0.11x

σ

n = 38

−= 1.82

z

0

The probability that the mean height of the 38 bushes is less than

7.8 feet is 0.0344.

= 0.0344

P ( < 7.8)x

17


Example:



that the mean score of 25 randomly selected students is between

75 and 79.


− −x

x

x µz

σ1

75 78= =

1.6−= 1.88

− −x µz

σ2

79 78= =

1.6= 0.63

0

z

?? 0.63−1.88

= 78

σ 8= = = 1.6

n 25

x

x

µ

σ

Continued.

P (75 < < 79)x

75 7978x


Example continued:


Approximately 70.56% of the 25 students will have a mean score

between 75 and 79.

= 0.7357 − 0.0301 = 0.7056

0

z

?? 0.63−1.88

P (75 < < 79)x

P(75 < < 79) = P(−1.88 < z < 0.63) = P(z < 0.63) − P(z < −1.88) x

75 7978x


Example:

The population mean salary for auto mechanics is µ =

$34,000 with a standard deviation of σ = $2,500. Find the probability

that the mean salary for a randomly selected sample of 50 mechanics

is greater than $35,000.

Probabilities of x and x

− −= x

x

x µz

σ35000 34000

=353.55

= 2.83

0z

?2.83

=

= 34000

2500= = 353.55

50

x

x

µ

σσ

n= P (z > 2.83) = 1 − P (z < 2.83)

= 1 − 0.9977 = 0.0023

The probability that the mean salary

for a randomly selected sample of 50

mechanics is greater than $35,000 is

0.0023.

3500034000

P ( > 35000)x

x

18


Example:

The population mean salary for auto mechanics is µ =

$34,000 with a standard deviation of σ = $2,500. Find the

probability that the salary for one randomly selected mechanic is

greater than $35,000.


- 35000 -34000= =

2500x µ

zσ

= 0.4

0z

?0.4

= 34000

= 2500

µ

σ= P (z > 0.4) = 1 − P (z < 0.4)

= 1 − 0.6554 = 0.3446

The probability that the salary for

one mechanic is greater than

$35,000 is 0.3446.

(Notice that the Central Limit Theorem does not apply.)

3500034000

P (x > 35000)

x


Example:

The probability that the salary for one randomly selected mechanic is

greater than $35,000 is 0.3446. In a group of 50 mechanics,

approximately how many would have a salary greater than $35,000?


P(x > 35000) = 0.3446This also means that 34.46% of mechanics

have a salary greater than $35,000.

You would expect about 17 mechanics out of the group of 50

to have a salary greater than $35,000.

34.46% of 50 = 0.3446 × 50 = 17.23

§ 5.5

Normal Approximations to

Binomial Distributions

19


Normal Approximation

The normal distribution is used to approximate the binomial

distribution when it would be impractical to use the binomial

distribution to find a probability.

Normal Approximation to a Binomial Distribution

If np ≥ 5 and nq ≥ 5, then the binomial random variable x is

approximately normally distributed with mean

and standard deviation

=σ npq .

=µ np


Normal Approximation

Example:

Decided whether the normal distribution to approximate x may be

used in the following examples.

1. Thirty-six percent of people in the United States own a dog.

You randomly select 25 people in the United States and ask

them if they own a dog.

2. Fourteen percent of people in the United States own a cat.

You randomly select 20 people in the United States and ask

them if they own a cat.

= (25)(0.36) = 9np

= (25)(0.64) = 16nq

Because np and nq are greater than 5, the

normal distribution may be used.

= (20)(0.14) = 2.8np

= (20)(0.86) = 17.2nq

Because np is not greater than 5, the normal

distribution may NOT be used.


Correction for Continuity

The binomial distribution is discrete and can be represented by a

probability histogram.

This is called the correction for continuity.

When using the continuous normal

distribution to approximate a binomial distribution, move 0.5

unit to the left and right of the midpoint to include all possible

x-values in the interval.

To calculate exact binomial probabilities, the

binomial formula is used for each value of x

and the results are added.

Exact binomial probability

cx

P(x = c)

P(c− 0.5 < x < c + 0.5)

Normal

approximation

cx

c + 0.5c − 0.5

20


Correction for Continuity

Example:

Use a correction for continuity to convert the binomial intervals to a

normal distribution interval.

1. The probability of getting between 125 and 145 successes,

inclusive.

The discrete midpoint values are 125, 126, …, 145.

The continuous interval is 124.5 < x < 145.5.

2. The probability of getting exactly 100 successes.The discrete midpoint value is 100.

The continuous interval is 99.5 < x < 100.5.

3. The probability of getting at least 67 successes.

The discrete midpoint values are 67, 68, ….

The continuous interval is x > 66.5.


Guidelines

Using the Normal Distribution to Approximate Binomial Probabilities

In Words In Symbols

1. Verify that the binomial distribution applies.

2. Determine if you can use the normal distribution to

approximate x, the binomial variable.

3. Find the mean µ and standard deviationσ for

the distribution.

4. Apply the appropriate continuity correction. Shade the

corresponding area under the normal curve.

5. Find the corresponding z-value(s).

6. Find the probability.

=σ n p q

-=

x µz

σ

Specify n, p, and q.

Is np ≥ 5?

Is nq ≥ 5?

=µ n p

Add or subtract 0.5 from

endpoints.

Use the Standard Normal

Table.


Approximating a Binomial Probability

Example:

Thirty-one percent of the seniors in a certain high school plan to attend

college. If 50 students are randomly selected, find the probability that less

than 14 students plan to attend college.

np = (50)(0.31) = 15.5

nq = (50)(0.69) = 34.5

The variable x is approximately normally

distributed with µ = np = 15.5 and

= (50 )(0 .31 )(0 .69 ) = 3 .27 .σ = n p q

P(x < 13.5)

Correction for

continuity

- --=

13 .5 15 .5= = 0 .61

3 .27x µ

zσ

= P(z < −0.61)

10 15

x

20

µ= 15.5

13.5

= 0.2709

The probability that less than 14 plan to attend college is 0.2079.

21


Approximating a Binomial Probability

Example:

A survey reports that forty-eight percent of US citizens own computers.

45 citizens are randomly selected and asked whether he or she owns a

computer. What is the probability that exactly 10 say yes?

np = (45)(0.48) = 12

nq = (45)(0.52) = 23.4 = = (45 )(0 .48 )(0 .52 ) = 3 .35σ n pq

P(9.5 < x < 10.5)

Correction for

continuity

5 10

x

15

µ = 12

9.5

= 0.0997

The probability that exactly 10 US

citizens own a computer is 0.0997.

= 12µ

10.5

= P(−0.75 < z − 0.45)

Chapter 5 Normal Probability Distributions Normal Probability Distributions Chapter 5 5.1...

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